If a, b, c are in G.P. then (1/a2 - b2) + (1/b2) is:

Question

If a, b, c are in G.P. then (1/a2 - b2) + (1/b2) is: (A) (1/c2 - b2) (B) (1/b2 - c2) (C) (1/c2 - a2) (D) (1/b2 - a2)

Tardigrade Admin · Accepted Answer

As given: a, b, c are in G.P. ⇒ b2 = ac The given expression: (1/a2 - b2) + (1/b2) = (1/a2 -ac) + (1/ac) [using b2 = ac ] = (1/a(a-c) ) + (1/ac) = (c+a-c/ac(a-c)) = (a/ac(a-c)) = (1/ac -c2) = (1/b2 -c2) [using ac = b2 ]

Q. If a, b, c are in G.P. then $\frac{1}{a ^{2} - b ^{2}} + \frac{1}{b ^{2}}$ is:

$\frac{1}{c ^{2} - b ^{2}}$

$\frac{1}{b ^{2} - c ^{2}}$

$\frac{1}{c ^{2} - a ^{2}}$

$\frac{1}{b ^{2} - a ^{2}}$

Q. If a, b, c are in G.P. then a2−b21​+b21​ is:

c2−b21​

b2−c21​

c2−a21​

b2−a21​

As given : a, b, c are in G.P. ⇒b2=ac The given expression : a2−b21​+b21​ =a2−ac1​+ac1​ [using b2=ac] =a(a−c)1​+ac1​=ac(a−c)c+a−c​ =ac(a−c)a​=ac−c21​=b2−c21​ [using ac=b2]

Q. If a, b, c are in G.P. then $\frac{1}{a ^{2} - b ^{2}} + \frac{1}{b ^{2}}$ is:

$\frac{1}{c ^{2} - b ^{2}}$

$\frac{1}{b ^{2} - c ^{2}}$

$\frac{1}{c ^{2} - a ^{2}}$

$\frac{1}{b ^{2} - a ^{2}}$